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Chain Rule of Differentiation

Prove that : ...

Question

Prove that : ${sec}^{6} θ = {tan}^{6} θ + 3 {tan}^{2} θ {sec}^{2} θ + 1$

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Solution

LHS $= {sec}^{6} θ$

$= {(1 + {tan}^{2} θ)}^{3}$

$= 1 + {tan}^{6} θ + 3 {tan}^{4} θ + 3 {tan}^{2} θ$

$= 1 + {tan}^{6} θ + 3 {tan}^{2} θ (1 + {tan}^{2} θ)$

$= 1 + {tan}^{6} θ + 3 {tan}^{2} θ {sec}^{2} θ$

$= R H S$

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Similar questions

{tan}^{6} θ + 3 {tan}^{4} θ + 3 {tan}^{2} = {sec}^{6} θ - 1

{sec}^{6} θ - {tan}^{6} θ = 1 + 3 {sec}^{2} θ {tan}^{2} θ

Q. Prove the identify :

{sec}^{2} θ ({sec}^{2} θ - 2) + 1 = {tan}^{4} θ .

{(csc θ + cot θ)}^{2} = \frac{sec θ + 1}{sec θ - 1}

(cosec θ + cot θ)^{2} = \frac{sec θ + 1}{sec θ - 1}

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